Year 2 Pure · Chapter 11

Parametric Equations

Specification: Edexcel Pure 3.3, 3.4 — Coordinate geometry · AQA section C3, C4

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In this chapter

1. Curves traced by a parameter
2. Finding the Cartesian equation
3. Intersections
4. Modelling with parametric equations
Practice

Key facts — this chapter

A parametric curve gives $x$ and $y$ separately in terms of a third variable, the parameter, often $t$ or $\theta$ .
Each value of the parameter gives one point $(x, y)$ . As the parameter changes, the point traces the curve.
To find the Cartesian equation, eliminate the parameter: make it the subject of one equation and substitute, or use an identity such as $\sin^2 t + \cos^2 t \equiv 1$ .
The values $x$ can take are the range of $x(t)$ , which may restrict the Cartesian curve to only part of its graph.
To find where a curve meets a line, substitute $x(t)$ and $y(t)$ into the line’s equation and solve for $t$ .

A ball in flight has a horizontal position and a height, and both change with time. Writing each in terms of time $t$ describes its path, and says where the ball is at every moment, which an equation linking $x$ and $y$ alone can’t do.

1. Curves traced by a parameter

Key result

A parametric curve is given by two equations

x = f(t), \qquad y = g(t)

with the parameter $t$ taking values in some interval. Each value of $t$ gives one point on the curve.

Worked example 11.1 · Plotting from a table

Sketch the curve $x = t^2$ , $y = 2t$ for $-2 \le t \le 2$ .

Work out a point for each of a few values of $t$ :

$t$	$-2$	$-1$	$0$	$1$	$2$
$x = t^2$	$4$	$1$	$0$	$1$	$4$
$y = 2t$	$-4$	$-2$	$0$	$2$	$4$

Plot the points in order of $t$ and join them with a smooth curve. It is a parabola on its side, traced from $(4, -4)$ up to $(4, 4)$ .

The parabola x = t squared, y = 2t, opening to the right, with the points for t = minus 2, minus 1, 0, 1 and 2 marked and labelled, and arrows showing the direction of increasing t from bottom to top — x = t², y = 2t, traced as t increases from −2 to 2.

Step the parameter t and watch the point (x, y) move. The solid part is the path traced so far; the faint dashed part is still to come.

2. Finding the Cartesian equation

To find an equation linking $x$ and $y$ alone, eliminate the parameter.

Worked example 11.2 · Eliminating the parameter

Find the Cartesian equation of the curve $x = t + 1$ , $y = t^2 - 3$ , for $-2 \le t \le 2$ , and state its domain.

Make $t$ the subject of the simpler equation: $t = x - 1$ .
Substitute into the other:
$y = (x - 1)^2 - 3$
As $t$ runs from $-2$ to $2$ , $x = t + 1$ runs from $-1$ to $3$ . So the curve is only the part of this parabola with $-1 \le x \le 3$ .

Worked example 11.3 · A circle

Show that the curve $x = 2 + 3\cos t$ , $y = -1 + 3\sin t$ , $0 \le t < 2\pi$ , is a circle, and give its centre and radius.

Isolate the trig functions: $\cos t = \tfrac{x - 2}{3}$ and $\sin t = \tfrac{y + 1}{3}$ .
Use $\cos^2 t + \sin^2 t \equiv 1$ :
$\begin{aligned} \frac{(x - 2)^2}{9} + \frac{(y + 1)^2}{9} &= 1 \\ (x - 2)^2 + (y + 1)^2 &= 9 \end{aligned}$

So it is the circle with centre $(2, -1)$ and radius $3$ .

Worked example 11.4 · A restricted curve

Find the Cartesian equation of the curve $x = \sin t$ , $y = \cos 2t$ , and state the values of $x$ for which it is defined.

Use the double-angle formula that contains only $\sin t$ :
$\begin{aligned} y = \cos 2t &= 1 - 2\sin^2 t \\ &= 1 - 2x^2 \end{aligned}$
$x = \sin t$ can only take values from $-1$ to $1$ . So the curve is the part of the parabola $y = 1 - 2x^2$ with $-1 \le x \le 1$ . It is traced backwards and forwards along this arc as $t$ increases.

Left: the parabola y = 1 minus 2 x squared drawn faintly, with only the arc from x = minus 1 to x = 1 drawn solid, which is the curve x = sin t, y = cos 2t. Right: the curve x = t squared minus 4, y = t cubed minus t, a loop that crosses itself at the point (minus 3, 0) — Left: x = sin t, y = cos 2t is only the arc of y = 1 − 2x² with −1 ≤ x ≤ 1. Right: x = t² − 4, y = t³ − t crosses itself at (−3, 0).

Common mistake

The Cartesian equation can describe more than the curve does. Always check which values $x$ and $y$ can take: in Example 11.4, $y = 1 - 2x^2$ goes on for ever, but the parametric curve stops at $x = \pm 1$ .

3. Intersections

To find where a parametric curve meets a line or another curve, substitute $x(t)$ and $y(t)$ into the other equation. That gives an equation in $t$ alone. Solve it, then turn each value of $t$ back into a point.

Worked example 11.5 · Meeting a line

Find the points where the curve $x = t^2$ , $y = 2t$ meets the line $y = x - 3$ .

Substitute into the line:
$\begin{aligned} 2t &= t^2 - 3 \\ t^2 - 2t - 3 &= 0 \\ (t - 3)(t + 1) &= 0 \end{aligned}$
$t = 3$ gives the point $(9, 6)$ , and $t = -1$ gives $(1, -2)$ .

Worked example 11.6 · Meeting the axes

The curve $C$ is $x = t^2 - 4$ , $y = t^3 - t$ . Find where $C$ meets the coordinate axes.

On the $y$ -axis, $x = 0$ , so $t^2 = 4$ and $t = \pm 2$ . Then $y = 8 - 2 = 6$ or $y = -8 + 2 = -6$ : the points $(0, 6)$ and $(0, -6)$ .
On the $x$ -axis, $y = 0$ , so $t(t^2 - 1) = 0$ and $t = 0$ or $\pm 1$ . Then $t = 0$ gives $(-4, 0)$ , and both $t = 1$ and $t = -1$ give $(-3, 0)$ .

Two values of $t$ give the same point, so the curve passes through $(-3, 0)$ twice: it crosses itself there, as the figure above shows.

4. Modelling with parametric equations

Worked example 11.7 · A projectile

A ball is kicked from the ground. After $t$ seconds it is $x = 20t$ metres along the ground and $y = 15t - 4.9t^2$ metres high. Find how far it travels before it lands, its greatest height, and the Cartesian equation of its path.

It lands when $y = 0$ : $t(15 - 4.9t) = 0$ , so $t = \tfrac{15}{4.9} = 3.061\ldots$ It is then $x = 20 \times 3.061\ldots = 61.2$ m away (3 s.f.).
$y$ is a quadratic in $t$ , greatest halfway between its roots, at $t = \tfrac{15}{9.8} = 1.530\ldots$ :
$\begin{aligned} y &= 15 \times 1.530\ldots \\ &\quad - 4.9 \times 1.530\ldots^2 \\ &= 11.5 \text{ m (3 s.f.)} \end{aligned}$
$t = \tfrac{x}{20}$ , so
$\begin{aligned} y &= 15 \times \tfrac{x}{20} - 4.9 \times \tfrac{x^2}{400} \\ &= 0.75x - 0.01225x^2 \end{aligned}$
for $0 \le x \le 61.2$ .

Practice

Twelve questions in three tiers. Try each one properly before opening the solution — the solutions show full method, not just answers.

Warm-up

1 [2 marks]

Find the Cartesian equation of the curve $x = 2t - 1$ , $y = t^2 + 1$ .

Show solution

$t = \tfrac{x + 1}{2}$ , so

y = \left(\frac{x + 1}{2}\right)^2 + 1 = \frac{(x + 1)^2}{4} + 1

2 [2 marks]

Describe the curve $x = 4\cos\theta$ , $y = 4\sin\theta$ , $0 \le \theta < 2\pi$ .

Show solution

$x^2 + y^2 = 16\cos^2\theta + 16\sin^2\theta = 16$ , so the curve is the circle with centre $(0, 0)$ and radius $4$ .

3 [2 marks]

A curve is given by $x = t^3$ , $y = t^2 - 1$ .

Find the point where $t = 2$ . [1]
Find the value of $t$ at the point $(-1, 0)$ . [1]

Show solution

(a) $x = 8$ and $y = 3$ , so the point is $(8, 3)$ .

(b) $t^3 = -1$ gives $t = -1$ , and then $y = 1 - 1 = 0$ as required.

4 [2 marks]

Find the Cartesian equation of the curve $x = \tfrac{1}{t}$ , $y = t + 1$ , where $t \ne 0$ .

Show solution

$t = \tfrac{1}{x}$ , so $y = \tfrac{1}{x} + 1$ .

Standard

5 [3 marks]

A curve is given by $x = 3 + 2\cos t$ , $y = -1 + 2\sin t$ , $0 \le t < 2\pi$ . Find its Cartesian equation, and describe the curve.

Show solution

$\cos t = \tfrac{x - 3}{2}$ and $\sin t = \tfrac{y + 1}{2}$ . Using $\cos^2 t + \sin^2 t \equiv 1$ :

(x - 3)^2 + (y + 1)^2 = 4

It is the circle with centre $(3, -1)$ and radius $2$ .

6 [4 marks]

A curve is given by $x = \cos t$ , $y = \sin 2t$ . Show that its Cartesian equation is $y^2 = 4x^2(1 - x^2)$ .

Show solution

$y = \sin 2t = 2\sin t\cos t$ . Squaring removes the sign of $\sin t$ :

\begin{aligned} y^2 &= 4\sin^2 t\cos^2 t \\ &= 4(1 - \cos^2 t)\cos^2 t \\ &= 4(1 - x^2)x^2 \end{aligned}

7 [4 marks]

A curve is given by $x = 2\sin t$ , $y = \cos 2t$ .

Find its Cartesian equation. [2]
State the values that $x$ and $y$ can take. [2]

Show solution

(a) $\cos 2t = 1 - 2\sin^2 t$ and $\sin t = \tfrac{x}{2}$ , so

y = 1 - 2 \times \frac{x^2}{4} = 1 - \frac{x^2}{2}

(b) $x = 2\sin t$ takes values $-2 \le x \le 2$ , and $y = \cos 2t$ takes values $-1 \le y \le 1$ .

8 [4 marks]

Find the points where the curve $x = t^2$ , $y = t^3$ meets the line $y = 2x$ .

Show solution

Substitute into the line:

\begin{aligned} t^3 &= 2t^2 \\ t^2(t - 2) &= 0 \end{aligned}

$t = 0$ gives $(0, 0)$ , and $t = 2$ gives $(4, 8)$ .

9 [4 marks]

Find the coordinates of the points where the curve $x = t^2 - 1$ , $y = t(t - 2)$ meets the coordinate axes.

Show solution

On the $y$ -axis, $x = 0$ , so $t = \pm 1$ . Then $t = 1$ gives $y = -1$ and $t = -1$ gives $y = 3$ : the points $(0, -1)$ and $(0, 3)$ .

On the $x$ -axis, $y = 0$ , so $t = 0$ or $t = 2$ . These give $(-1, 0)$ and $(3, 0)$ .

Exam style

10 [8 marks]

A ball is thrown from a window. Its position $t$ seconds later is modelled by $x = 12t$ and $y = 1.5 + 9t - 4.9t^2$ , where $x$ and $y$ are in metres, measured along the ground and up from the ground.

State the height from which the ball is thrown. [1]
Find the time at which the ball hits the ground, and how far it has then travelled horizontally, to 3 significant figures. [4]
Find the Cartesian equation of its path. [2]
Give one way in which the model could be improved. [1]

Show solution

(a) When $t = 0$ , $y = 1.5$ m.

(b) It hits the ground when $4.9t^2 - 9t - 1.5 = 0$ . By the quadratic formula, taking the positive root:

\begin{aligned} t &= \frac{9 + \sqrt{81 + 29.4}}{9.8} \\ &= 1.99 \text{ s (3 s.f.)} \end{aligned}

It has then travelled $x = 12 \times 1.990\ldots = 23.9$ m (3 s.f.).

(c) $t = \tfrac{x}{12}$ , so

\begin{aligned} y &= 1.5 + 9 \times \frac{x}{12} - 4.9 \times \frac{x^2}{144} \\ &= 1.5 + 0.75x - \frac{49x^2}{1440} \end{aligned}

(d) For example, include air resistance, which the model ignores.

11 [6 marks]

A curve is given by $x = \tan t$ , $y = \sec t$ , for $-\tfrac{\pi}{2} < t < \tfrac{\pi}{2}$ .

Show that the Cartesian equation is $y = \sqrt{1 + x^2}$ . [3]
Find the points where the curve meets the line $y = 2$ . [3]

Show solution

(a) $\sec^2 t = 1 + \tan^2 t$ , so $y^2 = 1 + x^2$ . For $-\tfrac{\pi}{2} < t < \tfrac{\pi}{2}$ , $\cos t > 0$ , so $y = \sec t > 0$ and $y = \sqrt{1 + x^2}$ .

(b) $\sec t = 2$ means $\cos t = \tfrac{1}{2}$ , so $t = \pm\tfrac{\pi}{3}$ . Then $x = \tan\left(\pm\tfrac{\pi}{3}\right) = \pm\sqrt3$ : the points are $\left(\sqrt3, 2\right)$ and $\left(-\sqrt3, 2\right)$ .

12 [7 marks]

The curve $C$ is given by $x = 1 + 2\cos t$ , $y = 2 + 2\sin t$ , $0 \le t < 2\pi$ . The line $y = x + 1$ meets $C$ at $A$ and $B$ .

Show that at $A$ and $B$ , $\tan t = 1$ . [2]
Find the exact coordinates of $A$ and $B$ . [3]
Show that $AB$ is a diameter of $C$ , and state its length. [2]

Show solution

(a) Substitute into the line: $2 + 2\sin t = 1 + 2\cos t + 1$ , so $\sin t = \cos t$ , and dividing by $\cos t$ gives $\tan t = 1$ .

(b) $t = \tfrac{\pi}{4}$ or $\tfrac{5\pi}{4}$ . With $\cos\tfrac{\pi}{4} = \sin\tfrac{\pi}{4} = \tfrac{\sqrt2}{2}$ , the points are $\left(1 + \sqrt2,\ 2 + \sqrt2\right)$ and $\left(1 - \sqrt2,\ 2 - \sqrt2\right)$ .

(c) $C$ is the circle with centre $(1, 2)$ and radius $2$ . The midpoint of $AB$ is $(1, 2)$ , the centre, so $AB$ is a diameter, of length $4$ .